In the given figure, if AB || DC then find ∠x, ∠y and ∠z.

1.	In the given figure, if AB \|\| DC then find ∠x, ∠y and ∠z.
Answer» In ∆BCE 88° = 22° + ∠z (by ext. angle property) ∠z = 88° – 22° ⇒ ∠z = 66° Also AB \|\| DC 88° + ∠y = 180° (interior angle property) ∠y = 180° – 88° ⇒ ∠y = 92° Again, 102° + ∠DAB = 180° (linear pair axiom) ∠DAB = 180° – 102° = 78° AB \|\| DC, then x + 78° = 180° (The sum of the interior angles on the same of a transversal is 180°) ⇒ x = 102° Hence ∠x = 102°, ∠y = 92°, ∠z = 66°

Answer»

In ∆BCE

88° = 22° + ∠z (by ext. angle property)

∠z = 88° – 22°

⇒ ∠z = 66°

Also AB || DC

88° + ∠y = 180° (interior angle property)

∠y = 180° – 88°

⇒ ∠y = 92°

Again, 102° + ∠DAB = 180° (linear pair axiom)

∠DAB = 180° – 102° = 78°

AB || DC, then x + 78° = 180° (The sum of the interior angles on the same of a transversal is 180°)

⇒ x = 102°

Hence ∠x = 102°, ∠y = 92°, ∠z = 66°

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